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Proof step 5 from proof two-loup-08 (previous step)

m + n = m0 * 2 + n1, m + 1 < m0, m < n0, n < m0, m < n, n < n0, m + 1 < n0, m < m0, n1 < 2, n0 ≤ m0 ⊦ m + n = 0, 2 = 0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m + n = m0 * 2 + n1, n < m0, m < n0, m < n, n < n0, m + 1 < n0, n1 < 2, m < m0, m + 1 < m0, n0 ≤ m0 ⊦ m + n = 0, m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic